of 1 17
The Meter, Second, and Kilogram As Natural Units In A Wave Theory Solution For The
Solar System
By
Ian Beardsley
Copyright © 2025!
of 2 17
Abstract……………………………………………….3
Data Used In This Paper………………………….…4
Introduction………………………………………….5
The Search For The Galactic Codex……………..…6
The Theory……………………………………………7
The Meter, Second, and Kilogram As
Natural Units In Our Theory………………………13
of 3 17
Abstract
A brief synopsis of the author’s theory for a wave solution of the Earth/Moon/Sun system that
incorporates his theory for the atom’s proton is presented, wherein it is found these are
described in terms of a characteristic time of one second. Then it is shown that in the context
of this theory that the meter, second, and kilogram may be natural units for length, mass, and
time, the basic units used to describe physical reality. As such it is necessary to the talk about
the origins of a second with the ancient Sumerians that settled down from crafting stone into
spearpoint to invent metallurgy, agriculture, writing, and mathematics; civilization in general.!
of 4 17
Data Used In This Paper
(Proton Mass)
(Planck Constant)
(Proton Radius)
(Gravitational Constant)
(light speed)
(Fine Structure Constant)
Earth day=(24)(60)(60)=86,400 seconds. Using the Moon’s orbital velocity at aphelion, and
Earth’s orbital velocity at perihelion we have:
m
P
: 1.67262 × 10
−27
kg
h : 6.62607 × 10
−34
J ⋅ s
r
p
: 0.833 × 10
−15
m
G:6.67408 × 10
−11
N
m
2
kg
2
c : 299,792,458m /s
α : 1/137
k
e
= 8.988E 9
Nm
2
C
2
K E
moon
=
1
2
(7.347673E 22kg)(966m /s)
2
= 3.428E 28J
K E
earth
=
1
2
(5.972E 24kg)(30,290m /s)
2
= 2.7396E 33J
of 5 17
Introduction One can speak of the structure of the long term structure of the solar system. The whole
object of developing a theory for the way planetary systems form is that they meet the following criterion:
They predict the Titius-Bode rule for the distribution of the planets; the distribution gives the planetary
orbital periods from Newton’s Universal Law of Gravitation. The distribution of the planets is chiefly
predicted by three factors: The inward forces of gravity from the parent star, the outward pressure
gradient from the stellar production of radiation, and the outward inertial forces as a cloud collapses into a
flat disc around the central star. These forces separate the flat disc into rings, agglomerations of material,
each ring from which a different planet forms at its central distance from the star (they have widths). In a
theory of planetary formation from a primordial disc, it should predict the Titius-Bode rule for the
distribution of planets today, which was the distribution of the rings from which the planets formed.
Also, the Earth has been in the habitable zone since 4 billion years ago when it was at 0.9 AU. Today it is
at 1AU, and that habitable zone can continue to 1.2 AU. So we can speak of this distance to the Earth over
much time. The Earth and Sun formed about 4.6 billion years ago. As the Sun very slowly loses mass over
millions of years as it burns fuel doing fusion, the Earth slips microscopically further out in its orbit over
long periods of time. The Earth orbit increases by about 0.015 meters per year. The Sun only loses
0.00007% of its mass annually. The Earth is at 1AU=1.496E11m. We have 0.015m/1.496E11m/
AU=1.00267E-13AU. So,
The Earth will only move out one ten thousandth of an AU in a billion years. Anatomically modern
humans have only been around for about three hundred thousand years. Civilization began only about six
thousand years ago.
The unit of a second becomes important in my theory. We got the second from the rotation period of the
Earth at the time the moon came to perfectly eclipse the Sun. The Moon slows the Earth rotation and this
in turn expands the Moon’s orbit, so it is getting larger, the Earth loses energy to the Moon. The Earth day
gets longer by 0.0067 hours per million years, and the Moon’s orbit gets 3.78 cm larger per year.
That is as the Earth’s day gets longer and the lunar orbit grows larger, we got the second at the time that
the Earth day was what it is during the epoch when the Moon perfectly eclipses the Sun, 24 hours.
The near perfect eclipse is a mystery in the sense that it came to happen when anatomically modern
humans arrived on the scene, even before that, perhaps around Homo Erectus and the beginning of the
Stone Age. The Earth day was 18 hours long, long before that, 1.4 billion years ago. Homo Erectus is
around two to three million years ago.
We find in our the theory that the characteristic time of 1 second of a wave solution for the Solar System
happens to be also the characteristic time for the Earth day of the 24 hours it is today, which was the same
when the ancient Sumerian formulated the second. We also find the characteristic time of the proton is 1
second, yielding its radius, and showing a mirroring of the microcosmos (subatomic particles) with the
macrocosmos (planetary systems).
(1.00267E − 13A U/year)(1E 9years) = 0.0001AU
of 6 17
The Search For The Galactic Codex
I have found that our Solar System has a fascinating mathematical structure underlying it. Our
planet is an extraordinary example of a life bearing world. The mathematical structure I have
found in my wave solution of the Solar system for the Earth/Moon/System could be taken as
characteristic of a star like our Sun. Naturally, in discovering something about our solar system
we would wonder if other star systems have a dynamic mathematical construction as well. As
such, we would want to go to other star systems and survey their characteristics, and any
ancient history of how other civilizations measured time and made calendars based on their
observations of celestial motions like we did with ours. We might guess that other civilizations
in the universe might discover such incredible mathematical structure underlying their planetary
systems as well and would, once they could achieve interstellar travel, begin a survey of the
mathematical structure behind star systems of other life bearing worlds and create a collection
of them done throughout the galaxy, and they may have worked with other civilizations
throughout the galaxy and compiled, if you will, a Galactic Codex. Some may even have
achieved intergalactic travel and found the thumbprints not just characteristic of star-types but
of galaxy-types. Let us look at what I have found our first entries could be in a galactic codex,
which would be for the star system we know best, our Solar System. We begin with the
characteristic time for our solution is one second and is given by the mass of the moon, ,
cubed:"
"
Yes, it does happen to be one second, our base unit of time we have today, ultimately given to
us by the ancient Sumerians when they invented mathematics thousands of years ago. This
becomes important, as we will see. The other extraordinary thing we find is that the Planck-
type constant for our solar system is given by the kinetic energy of the Earth, the third planet
where life is extraordinarily abundant, multiplied by one second."
"
We find life occurs so successfully when the Earth day is about what it is today (24 hours long)
giving a characteristic time of close to one second in terms of the kinetic energy of the Moon
and the Earth:"
"
I also find that this characteristic time of one second is characteristic of the proton, the most
fundamental unit that makes up matter, predicting its radius:"
"
"
M
m
ℏ
2
⊙
GM
3
m
⋅
1
c
= 1secon d
ℏ
⊙
= (1secon d )K E
e
KE
m
KE
e
(Ear thDay) = 1.1 − 1.3seconds
(
1
6α
2
4πh
Gc
)
⋅
r
p
m
p
= 1secon d
2
3
⋅
πr
p
α
4
Gm
3
p
1
3
⋅
h
c
= 1secon d
of 7 17
Where and are the radius and mass of a proton. Equating these two gives about the
radius of a proton:"
"
The ancient Sumerians were responsible for giving us the unit of a second because they
divided the earth’s rotation period, its day, into 24 hours, and the ancient Babylonians divided
each hour into 60 minutes, and each minute into 60 seconds, from the ancient Sumerians base
60 mathematics. Perhaps the most exciting entry in our galactic codex is:"
"
Where is the rotational angular momentum of the Earth. This specifies not only is the
rotation period of the Earth best measured by dividing the day into 24 units and 60 units, but
that such an optimization includes the mass and radius of the Earth. Another exciting entry in
our galactic codex is that during the time in the Earth’s history when the day is about 24 hours
which specifies close to a second from the kinetic energies of the Earth and Moon, the Moon
perfectly eclipsing the Sun as seen from the Earth, holds:"
"
is the orbital radius of the Earth, is the orbital radius of the Moon, is the
radius of the Sun, and is the radius of the Moon. There is a lot more that we will find in
the course of this paper regarding the exciting entries to be made in this Galactic Codex. We
will even find, with astronomy being what it is today, that we can begin to make entries in the
codex for other star systems. But of course, to really understand such star systems, we want
to go to them, and survey them not just physically, but archaeologically."
Perhaps, in our radio astronomy search for extraterrestrial intelligence (SETI) one of the
transmissions we might receive might be not just the physical characteristics for their star and
planet, but a Galactic Codex for many star systems. We even may be able to find traces of a
galactic codex here on earth now, left in the ruins of archaeological sites. Such examples could
be in clay Sumerian cuneiform tablets or in the megalithic yard which was perhaps a standard
length used to construct megalithic sites, like Stonehenge. We will look at that, too, in this
paper. "
The Theory
We begin with the characteristic time for our solution is one second and is given by the mass of the moon,
, cubed:
1.
r
p
m
p
r
p
=
2
3
⋅
h
cm
p
L
earth
ℏ
⊙
24 = 60
L
earth
r
planet
r
moon
=
R
star
R
moon
r
planet
r
moon
R
star
R
moon
M
m
ℏ
2
⊙
GM
3
m
⋅
1
c
= 1secon d
of 8 17
Where is a Planck-type constant for the Soar System. Yes, it does happen to be one second, our base
unit of time we have today, ultimately given to us by the ancient Sumerians when they invented
mathematics thousands of years ago. This becomes important, as we will see. The other extraordinary
thing we find is that the Planck-type constant for our solar system is given by the kinetic energy of the
Earth, the third planet where life is extraordinarily abundant, multiplied by one second.
2.
We find our equations hold so well when the Earth day is about what it is today (24 hours long) giving a
characteristic time of close to one second in terms of the kinetic energy of the Moon and the Earth:
3.
I also find that this characteristic time of one second is characteristic of the proton, the most fundamental
unit that makes up matter, predicting its radius:
4.
5.
Where and are the radius and mass of a proton. Equating these two gives about the radius of a
proton:
6.
This is very close to the value upon which the proton radius converged historically by two independent
methods which was 0.877E-15m. The result from our theory is
7.
The 0.877fm was challenged in 2010 by a third experiment making it 4% smaller and was 0.842E-15m. I
suggest it may be that the radius of a proton is actually
8.
utilizing an optimization characteristic of the golden ratio, as Nature often does, where is the
golden ratio. The most recent value is 0.833E-15m. In this case we say
ℏ
⊙
ℏ
⊙
ℏ
⊙
= (1secon d )K E
e
K E
m
K E
e
(Ear th Da y) = 1.1 − 1.3secon d s
(
1
6 α
2
4πh
G c
)
⋅
r
p
m
p
= 1secon d
2
3
⋅
π r
p
α
4
G m
3
p
1
3
⋅
h
c
= 1secon d
r
p
m
p
r
p
=
2
3
⋅
h
cm
p
r
p
=
2
3
⋅
6.62607E − 34
(299,792,458)(1.67262E − 27)
= 0.88094E − 15m
r
p
= ϕ ⋅
h
cm
p
= 0.816632E − 15m
ϕ = 0.618
of 9 17
9.
Where equation 5
is a little over a second, and 9
is a little under a second. The proton doesn’t have a precise size, but is fuzzy like a cloud of subatomic
particles. It can have a functional radius depending on the Nature of the problem, which determines the
definition of what its radius is.
The ancient Sumerians were responsible for giving us the unit of a second because they divided the
earth’s rotation period, its day, into 24 hours, and the ancient Babylonians divided each hour into 60
minutes, and each minute into 60 seconds from the ancient Sumerians base 60 mathematics. Perhaps the
most exciting equation in our theory is:
10.
Where is the rotational angular momentum of the Earth. This specifies not only that the rotation
period of the Earth is best measured by dividing the day into 24 units and 60 units, but that such an
optimization includes the mass and radius of the Earth. Another exciting equation in our theory is that
during the time in the Earth’s history when the day is about 24 hours which specifies close to a second
from the kinetic energies of the Earth and Moon, the Moon perfectly eclipsing the Sun as seen from the
Earth, holds:
11.
is the orbital radius of the Earth, is the orbital radius of the Moon, is the radius of the
Sun, and is the radius of the Moon. The Moon makes life so successful on Earth because it holds
the Earth at its inclination to its orbit around the Sun, preventing temperature extremes and allowing for
the seasons.
The Schrödinger Wave Equation must be solved to determine the energies and orbitals of the electron in
the hydrogen atom. In spherical coordinates it is
ϕ ⋅
π r
p
α
4
G m
3
p
1
3
⋅
h
c
= 1secon d
2
3
⋅
π r
p
α
4
G m
3
p
1
3
⋅
h
c
= 1secon d
ϕ ⋅
π r
p
α
4
G m
3
p
1
3
⋅
h
c
= 1secon d
L
earth
ℏ
⊙
24 = 60
L
earth
r
planet
r
moon
=
R
star
R
moon
r
planet
r
moon
R
star
R
moon
of 10 17
12.
It has the solutions
13.
14.
I find the solutions are for the Earth orbiting the Sun are:
15.
16.
is the solar radius, that of the Moon. For Earth , third planet. Because these have accuracies
close to 100% accurate we know that our equation for the solar system Planck constant is correct,
equation 2:
Let us now compute the value of and show that the units work…
Where from equation 5
Where is the radius of a proton, is the mass of a proton, is the speed of light, and is the fine
structure constant. We found this gives the characteristic time of one second in terms of a proton. We
guess the planetary scale is connected to the proton scale because the planets formed from the
protoplanetary disc and it is made of different combinations of protons. We derive the value of our solar
Planck constant:
=
−
ℏ
2
2 m
[
1
r
2
∂
∂r
(
r
2
∂
∂r
)
+
1
r
2
sin θ
∂
∂θ
(
sin θ
∂
∂θ
)
+
1
r
2
sin
2
θ
∂
2
∂ ϕ
2
]
ψ + V(r)ψ = E ψ
E
n
= −
Z
2
(k
e
e
2
)
2
m
e
2ℏ
2
n
2
r
n
=
n
2
ℏ
2
Z k
e
e
2
m
e
K E
e
= n
R
⊙
R
m
⋅
G
2
M
2
e
M
3
m
2ℏ
2
⊙
r
n
=
2ℏ
2
⊙
GM
3
m
⋅
R
⊙
R
m
⋅
1
n
R
⊙
R
m
n = 3
ℏ
⊙
= (1secon d )K E
e
ℏ
⊙
ℏ
⊙
= (hC )K E
e
hC = 1secon d
C =
1
3
⋅
1
α
2
c
2
3
⋅
π r
p
G m
3
p
r
p
m
p
c
α
C =
1
3
⋅
1
α
2
c
1
3
⋅
2 π r
p
G m
3
p
of 11 17
=
=
=
17.
I guessed since one second comes from the ancient Sumerians dividing the Earth day (rotation period)
into 24 hours, and those into 60 minutes, and those into 60 seconds, that this has to do with the rotational
angular momentum of the Earth, . I found, equation 10,…
.
Where
18.
The value is 2.5 which by modeling our Solar System is found to be the exponent in the pressure gradient
for the protoplanetary disc from which our Solar System formed. That is I found
19.
the pressure of the disc as a function of radius. Which suggests that the structure of the protoplanetary
disc could be governed by the same fundamental time of one second in the Earth’s rotation and that the
Earth’s formation process may be encoded in the same number we developed since ancient times to
describe time (24, 60). This is the solution to:
1
3
⋅
18769
299792458
1
3
⋅
2 π (0.833E − 15)
(6.67408E − 11)(1.67262E − 27)
3
1.55976565E 33
s
m
m
kg
3
⋅
s
2
kg
m
3
=
s
m
s
2
kg
2
m
2
=
s
m
⋅
s
kg ⋅ m
=
1
kg
⋅
s
2
m
2
1
C
= kg
m
2
s
2
=
1
2
mv
2
= en erg y
hC = (6.62607E − 34)(1.55976565E 33) = 1.03351secon d s ≈ 1.0secon d s
hC =
(
kg
m
s
2
⋅ m ⋅ s
)
(
1
kg
⋅
s
2
m
2
)
(
kg
m
2
s
)(
1
kg
⋅
s
2
m
2
)
= secon d s
K E
earth
=
1
2
(5.972E 24kg)(30,290m /s)
2
= 2.7396E 33J
ℏ
⊙
= (hC )K E
earth
= (1.03351s)(2.7396 E 33J ) = 2.8314E 33J ⋅ s
L
earth
L
earth
ℏ
⊙
24 = 60
L
earth
=
4
5
π M
e
f
e
R
2
e
P(R) = P
0
(
R
R
0
)
−
L
ear th
ℏ
⊙
of 12 17
20.
The protoplanetary disc that evolves into the planets has two forces that balance its pressure, the
centripetal force of the gas disc due to its rotation around the protostar and the inward gravitational
force on the disc from the protostar , and these are related by the density of the gas that makes
up the disc. It is the pressure gradient of the disc in radial equilibrium balancing the inward gravity and
outward centripetal force. In order to apply this to other star systems, we have to be able to predict the
radius of the habitable planet, presumably in the n=3 orbit. I found the answer to be in the Vedic literature
of India. They noticed that the diameter of the Sun is about 108 times the diameter of the Earth and that
the average distance from the Sun to the Earth is about 108 solar diameters, with 108 being a significant
number in Yoga. So I wrote the equivalent:
21.
The surprising result I found was, after applying it to the stars of many spectral types, with their different
radii and luminosities (the luminosities determine , the distances to the habitable zones) that the
radius of the planet always came out about the same, about the radius of the Earth. This may suggest
optimally habitable planets are not just a function of the distance from the star, which determines their
temperature, but are functions of their size and gravity probably because it is good for life chemistry. Here
are just a few examples using the data for several spectral types:
F8V Star
Mass: 1.18
Radius: 1.221
Luminosity: 1.95
F9V Star
Mass: 1.13
Radius: 1.167
Luminosity: 1.66
d P
dr
= − ρ
(
GM
⋆
r
2
−
v
2
ϕ
r
)
v
2
ϕ
/r
GM
⋆
/r
2
ρ
R
planet
= 2
R
2
⋆
r
planet
r
planet
M
⋆
= 1.18(1.9891E 30k g) = 2.347E 30kg
R
⋆
= 1.221(6.9634E 8m) = 8.5023E 8m
r
p
= 1.95L
⊙
AU = 1.3964AU(1.496E11m /AU ) = 2.08905E11m
R
p
=
2R
2
⋆
r
p
= 2
(8.5023E 8m)
2
2.08905E11m
=
6.92076E6m
6.378E6m
= 1.0851Ear th Ra dii
M
⋆
= 1.13(1.9891E 30k g) = 2.247683E 30kg
R
⋆
= 1.167(6.9634E 8m) = 8.1262878E 8m
r
p
= 1.66L
⊙
AU = 1.28841AU(1.496E11m /AU ) = 1.92746E11m
R
p
=
2R
2
⋆
r
p
= 2
(8.1262878E 8m)
2
1.92746E11m
=
6.852184E6m
6.378E6m
= 1.0743468Ear th R a dii
of 13 17
G0V Star
Mass: 1.06
Radius: 1.100
Luminosity: 1.35
G1V Star
Mass: 1.03
Radius: 1.060
Luminosity: 1.20
As you can see we consistently get about 1 Earth radius for the radius of every planet in the habitable
zone of each type of star. It might be that radius is right for life in terms of gravity and densities for the
elements. I got these results for the stars from spectral types F5V to K3V.
In order to get , the distance of the habitable planet from the star, we use the inverse square law for
luminosity of the star. If the Earth is in the habitable zone, and if the star is one hundred times brighter
than the Sun, then by the inverse square law the distance to the habitable zone of the planet is 10 times
that of what the Earth is from the Sun. Thus we have in astronomical units the habitable zone of a star is
given by:
22.
The Meter, Second, and Kilogram As Natural Units In Our Theory
In that the characteristic times for our solutions of the Earth/Moon/Sun System and the proton
are 1 second we can say that the second is a natural unit of time. We have"
M
⋆
= 1.06(1.9891E 30kg) = 2.108446E 30k g
R
⋆
= 1.100(6.9634E 8m) = 7.65974E 8m
r
p
= 1.35L
⊙
AU = 1.161895AU(1.496E11m /AU ) = 1.7382E11m
R
p
=
2R
2
⋆
r
p
= 2
7.65974E 8m)
2
1.7382E11m
=
6.751E6m
6.378E6m
= 1.05848Ear th R a dii
M
⋆
= 1.03(1.9891E 30k g) = 2.11E 30kg
R
⋆
= 1.060(6.9634E 8m) = 7.381E 8m
r
p
= 1.20L
⊙
AU = 1.0954AU(1.496E11m /AU ) = 1.63878589E11m
R
p
=
2R
2
⋆
r
p
= 2
7.3812E 8m)
2
1.63878589E11m
=
6.6491E6m
6.378E6m
= 1.0425Ear th Ra dii
r
planet
r
planet
=
L
⋆
L
⊙
AU
ℏ
2
⊙
GM
3
m
⋅
1
c
= 1secon d
ℏ
⊙
= (1secon d )K E
e
KE
m
KE
e
(Ear thDay) = 1.1 − 1.3seconds
of 14 17
We now want to proceed to show that the meter is a natural unit of length, and that the
kilogram is a natural unit of mass even though they are human inventions that were not derived
from physical properties of the Solar System or proton. To show this is the case for a meter, we
notice that the period of a pendulum swinging under earth gravity with length of 1 meter has a
period of about 2 seconds, meaning a swing left, or right, is about 1 second. We have"
23. "
Which means"
24. "
The surface gravity for the Earth is in terms of its mass and radius:"
25. "
26. "
The Length of a pendulum, L, then, is about 1 meter:"
27. "
= "
= "
This equation results in a quite nice equation. We write meters per second squared in terms of
it and equation 1, which is:"
(
1
6α
2
4πh
Gc
)
⋅
r
p
m
p
= 1secon d
ϕ ⋅
πr
p
α
4
Gm
3
p
1
3
⋅
h
c
= 1secon d
T = 2π
L
g
1secon d = π
L
g
g =
GM
e
R
2
e
L =
(1sec)
2
GM
e
π
2
R
2
e
L =
ℏ
4
⊙
G
⋅
M
e
M
6
m
⋅
1
R
2
e
c
2
⋅
1
π
2
(2.8314E33J ⋅ s)
4
(5.972E 24kg)
(6.674E − 11)(7.347673E 22kg)
6
⋅
1
(6.378E6m)
2
(299792458m /s)
2
⋅
1
π
2
1.01281meters ≈ 1.0meters
of 15 17
"
"
And, we notice that the surface gravity on the Moon is about about meters per second
squared, where is the golden ratio:"
28. "
The result is:"
29. "
Now we want to show that the kilogram may be a natural unit of mass. Natural because it is an
intermediate mass between say a proton and the Earth. We do this by taking the geometric
mean between the two and introduce the ratio of the lunar mass and that and the Earth:"
30. "
= "
Thus we have the basic physical quantities that describe the Universe: mass, length, and time
as natural in our theory if written in meters, seconds, and kilograms:"
"
"
"
The kilogram is also about the geometric mean between the proton and the planet Saturn."
L =
(1sec)
2
GM
e
π
2
R
2
e
ℏ
2
⊙
GM
3
m
⋅
1
c
= 1secon d
Φ
Φ = 1.618
g
moon
=
GM
m
R
2
m
= Φ
m
s
2
R
2
e
R
2
m
=
Φ
π
2
M
e
M
m
M = m
p
M
e
M
m
M
e
(1.67262E − 27kg)
5.972E 24kg
7.347573E 22kg
5.972E 25kg = 0.9010kg ≈ 1k ilogram
ℏ
2
⊙
GM
3
m
⋅
1
c
= 1secon d
L =
ℏ
4
⊙
G
⋅
M
e
M
6
m
⋅
1
R
2
e
c
2
⋅
1
π
2
= 1meter
M = m
p
M
e
M
m
M
e
= 1kilogra m
of 16 17
"
"
of 17 17
The Author"